
Step 1: Magnetic Field Due to the Arc
For the arc with radius \( a \) and angle \( \frac{3\pi}{2} \), the magnetic field at the origin is: \[ B_1 = \frac{\mu_0 I}{4\pi a} \]
Step 2: Magnetic Field Due to the Straight Segment
For the straight segment of the wire: \[ B_2 = \frac{\mu_0 I}{4\pi a} \left( \frac{3\pi}{2} \right) \]
Step 3: Magnetic Field Due to Other Segments
Since the magnetic field due to the straight segments at the origin is zero: \[ B_3 = 0 \]
Step 4: Calculate the Total Magnetic Field
Thus, the total magnetic field at the origin is: \[ B = \frac{\mu_0 I}{4\pi a} \left( \frac{3\pi}{2} \right) \]
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____. 
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.